Why are all Rényi entropies equal for Clifford dynamics?

Question

In this paper, by Adam Nahum et al., the authors trivially states that "For Clifford dynamics all Rényi entropies are equal ... " which is not trivial to me.

Is there a paper or lecture notes on why this is true or a good explanation?

Answer 1

Let me report here a brief demonstration that I used to convince myself. The R'enyi entropies are \begin{equation*} S_\mathrm{A}^\alpha = \frac{1}{1-\alpha} \log_2 \text{Tr} (\rho_\mathrm{A}^{\alpha}) \end{equation*} In a nutshell, this is a consequence of the reduced density matrix $\rho_{\mathrm{A}}$ being a stabilizer itself, and hence it can be written a sum over the elements of the stabilizers group $\mathcal{S}_{\mathrm{A}}$. Here $\mathcal{S}_{\mathrm{A}}$ indicates the stabilizer group of all stabilizers that act trivially (identity) on the B part as defined in Appendix C of this paper. We begin with \begin{equation} \rho_\mathrm{A} = \frac{|\mathcal{S}_\mathrm{A}|}{2^{|\mathrm{A}|}} \frac{1}{|\mathcal{S}_{\mathrm{A}}|} \sum_{s\in \mathcal{S}_\mathrm{A}}s. \end{equation} By squaring the density matrix, we obtain: \begin{equation} \rho_{\mathrm{A}}^2 = \frac{1}{2^{2 |\mathrm{A}|}} \sum_{s \in \mathcal{S}_{\mathrm{A}}} \sum_{s' \in \mathcal{S}_{\mathrm{A}}} s s' = \frac{1}{2^{2 |\mathrm{A}|}} |\mathcal{S}_{\mathrm{A}}|\sum_{s \in \mathcal{S}_{\mathrm{A}}} s = \frac{|\mathcal{S}_{\mathrm{A}}|}{2^{|\mathrm{A}|}} \rho_{\mathrm{A}} \end{equation} In the second equality we utilize the fact $\mathcal{S}_{\mathrm{A}}$ is a group, hence the product between its elements generate elements within the same group, and this happens exactly $|\mathcal{S}_{\mathrm{A}}|$ times. It is evident that in general $\rho_{\mathrm{A}}^{\alpha} = (|\mathcal{S}_{\mathrm{A}}| / 2^{|\mathrm{A}|})^{\alpha - 1} \rho_{\mathrm{A}}$. By substituting this into the expression for the R'enyi entropy, we obtain \begin{align*} S_{\mathrm{A}}^{\alpha} & = \frac{1}{1-\alpha} \log_2 \text{Tr} \bigg[ \bigg(\frac{|\mathcal{S}|}{ 2^{|\mathrm{A}|}} \bigg)^{\alpha - 1} \rho_{\mathrm{\alpha}} \bigg] \\ & = \frac{1}{1-\alpha} \log_2 \bigg(\frac{|\mathcal{S}_\mathrm{A}}{ 2^{|\mathrm{A}|}} \bigg)^{\alpha - 1} \text{Tr} \rho_{\mathrm{A}} \\ & = |\mathrm{A}| - \log_2|\mathcal{S}_{\mathrm{A}}| \\ \end{align*} which is exactly the result we achieved in this paper and similar as this paper.

Answer 2

In stabilizer states (which are the states produced by clifford dynamics) the entanglement spectrum is flat, i.e., all the non-zero Schmidt coefficients are equal. Starting from there, it is straightforwards to check that all Renyi entropies are equal, namely $\log D$, where $D$ is the number of non-zero Schmidt coefficients.